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3 years ago

Find the length of side x in simplest radical form with a rational denominator.

Mathematics

1 answer:

madreJ [45]3 years ago

8 0

Answer:

x=2√5

Step-by-step explanation:

1st method:

sin(30)=opposite/hypotenuse=√5/x

sin(30)x=√5

x=√5/sin(30)

x=2√5

2nd method:

cos(60)=adjacent/hypotenuse=√5/x

cos(60)x=√5

x=√5/cos(60)

x=2√5

3rd method:

This is a 30-60-90 triangle, so we can multiply the short side, √5, by 2, to get x=2√5

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A development economist is studying income growth in a rural area of a developing country. The last census of the population of

Anna [14]

Answer:

Reject H0 if x > cv

Step-by-step explanation:

The hypothesis :

H0 : μ ≤ 425

H1 : μ > 425

Standard deviation, s = √2500 = 50

Sample size, n = 100

xbar = 433.5

The test statistic, Z :

(xbar - μ) / s/√n

(433.5 - 425) / 50/10

Z = 8.5 / 5 = 1.7

The decision region ;

|Z| > Z0.01 ; Reject H0

From Z table ;critical value, Z0.01 = 2.33

1.7 < 2.33 ; We fail to reject then Null and conclude that thee is no significant evidence support the claim that population mean income has increased.

7 0

2 years ago

Need help ASAP!! I'll give brainliest to the correct answer!!!

Scilla [17]

Answer:

B. 38.5

Step-by-step explanation:

Not 100% sure, but

41+36=77

77/2=

=38.5

4 0

3 years ago

-5/6x-7/30x+1/5x-52 please help me guys cuz imk not good with math

Marina CMI [18]

Answer:

$\frac{-13}{15} x-52$

Step-by-step explanation:

$\frac{-5}{6}x-\frac{7}{30}x+\frac{1}{5}x-52$

$=\frac{-5}{6}x+\frac{-7}{30}x+\frac{1}{5}x+-52$

Combine Like Terms:

$=\frac{-5}{6}x+\frac{-7}{30}x+\frac{1}{5}x+-52$

$=(\frac{-5}{6}x+\frac{-7}{30}x+\frac{1}{5}x)+(-52)$

$=\frac{-13}{15}x+-52$

Therefore, $\frac{-13}{15}x+-52$ will be your final answer.

7 0

2 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago

Two integers from 1 through 40 are chosen by a random number generator. write your answer as a fraction in simplest form a/b. p(

jasenka [17]

Probability is 1/40.

5 0

3 years ago